Question

If the normal to the curve $y=f\left(x\right)$ at the point $(3,4)$ makes $\frac{3\mathrm{\pi }}{4}$ with the positive $x-$axis then $f\text{'}\left(3\right)$ $=?$

• A. $-1$
• B. $\frac{-3}{4}$
• C. $\frac{4}{3}$
• D. $1$

Answer 1

The correct option is D

$1$

Explanation for the correct answer:

The normal makes an angle of $\frac{3\mathrm{\pi }}{4}$ with the positive $x-$axis.

Let the slope of the normal be $m_n$

$\Rightarrow$$m_n=\tan \frac{3\mathrm{\pi }}{4}$

$=\tan \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right)$

$=-\tan \left(\frac{\mathrm{\pi }}{4}\right)$

$\Rightarrow$$m_n=-1$$...\left(i\right)$

The first order derivative of the equation of the curve at a point gives the slope of the tangent to the curve at that point.

Let $m_t$ be the slope of the tangent to the curve at point $(3,4)$

${\frac{dy}{dx}}_{\left(3,4\right)}=f\text{'}\left(x\right)=m_t$

$\Rightarrow$ $m_t=f\text{'}\left(3\right)$$...\left(ii\right)$

The tangent and the normal are perpendicular to each other. Hence, the product of their slopes is $-1$.

$\Rightarrow$ $m_t\times m_n=-1$

From $\left(i\right),\left(ii\right)$ we get

$\Rightarrow$$f\text{'}\left(3\right)\times \left(-1\right)=-1$

$\Rightarrow$ $f\text{'}\left(3\right)=1$

So, option $\left(D\right)$ is the correct answer.